272183
When an electric dipole $\vec{P}$ is placed in a uniform electric field $\vec{E}$ then at what angle between $\vec{P}$ and $\vec{E}$ the value of torque will be maximum?
1 ${{90}^{\circ }}$
2 ${{0}^{\circ }}$
3 ${{180}^{\circ }}$
4 ${{45}^{\circ }}$
Explanation:
(a)
NCERT Page-28 / N27
Electric Charges and Fields
272184
An electric dipole has a pair of equal and opposite point charges $q$ and $-q$ separated by a distance $2x$. The axis of the dipole is
1 from positive charge to negative charge
2 from negative charge to positive charge
3 perpendicular to the line joining the two charges drawn at the centre and pointing upward direction
4 perpendicular to the line joining the two charges drawn at the centre and pointing downward direction
Explanation:
(b)
NCERT Page-28 / N-27
Electric Charges and Fields
272185
The electric intensity due to a dipole of length $10~cm$ and having a charge of $500\mu C$, at a point on the axis at a distance $20~cm$ from one of the charges in air, is
1 $6.25\times {{10}^{7}}~N/C$
2 $9.28\times {{10}^{7}}~N/C$
3 $13.1\times {{10}^{11}}~N/C$
4 $20.5\times {{10}^{7}}~N/C$
Explanation:
(a) Given : Length of the dipole $\left( 2l \right)=10~cm=0.1~m$ or $l=$ $0.05~m$
Charge on the dipole $\left( q \right)=500\mu C=500\times {{10}^{-6}}C$ and distance of the point on the axis from the mid-point of the dipole $\left( r \right)=20+5=25~cm=0.25~m$.
We know that the electric field intensity due to dipole on the given point $\left( E \right)$
$=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2\left( q.2l \right)r}{{{\left( {{r}^{2}}-{{l}^{2}} \right)}^{2}}}=9\times {{10}^{9}}\times \frac{2\left( 500\times {{10}^{-6}}\times 0.1 \right)\times 0.25}{{{\left[ {{(0.25)}^{2}}-{{(0.05)}^{2}} \right]}^{2}}}$
$=6.25\times {{10}^{7}}~N/C(k=1$ for air $)$
NCERT Page-28 / N-24
Electric Charges and Fields
272186
Intensity of an electric field $\left( E \right)$ depends on distance $r$, due to a dipole, is related as
1 $E\propto \frac{1}{r}$
2 $E\propto \frac{1}{{{r}^{2}}}$
3 $E\propto \frac{1}{{{r}^{3}}}$
4 $E\propto \frac{1}{{{r}^{4}}}$
Explanation:
(c) Intensity of electric field due to a Dipole
$E=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}\sqrt{3co{{s}^{2}}\theta +1}\Rightarrow E\propto \frac{1}{{{r}^{3}}}$
272183
When an electric dipole $\vec{P}$ is placed in a uniform electric field $\vec{E}$ then at what angle between $\vec{P}$ and $\vec{E}$ the value of torque will be maximum?
1 ${{90}^{\circ }}$
2 ${{0}^{\circ }}$
3 ${{180}^{\circ }}$
4 ${{45}^{\circ }}$
Explanation:
(a)
NCERT Page-28 / N27
Electric Charges and Fields
272184
An electric dipole has a pair of equal and opposite point charges $q$ and $-q$ separated by a distance $2x$. The axis of the dipole is
1 from positive charge to negative charge
2 from negative charge to positive charge
3 perpendicular to the line joining the two charges drawn at the centre and pointing upward direction
4 perpendicular to the line joining the two charges drawn at the centre and pointing downward direction
Explanation:
(b)
NCERT Page-28 / N-27
Electric Charges and Fields
272185
The electric intensity due to a dipole of length $10~cm$ and having a charge of $500\mu C$, at a point on the axis at a distance $20~cm$ from one of the charges in air, is
1 $6.25\times {{10}^{7}}~N/C$
2 $9.28\times {{10}^{7}}~N/C$
3 $13.1\times {{10}^{11}}~N/C$
4 $20.5\times {{10}^{7}}~N/C$
Explanation:
(a) Given : Length of the dipole $\left( 2l \right)=10~cm=0.1~m$ or $l=$ $0.05~m$
Charge on the dipole $\left( q \right)=500\mu C=500\times {{10}^{-6}}C$ and distance of the point on the axis from the mid-point of the dipole $\left( r \right)=20+5=25~cm=0.25~m$.
We know that the electric field intensity due to dipole on the given point $\left( E \right)$
$=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2\left( q.2l \right)r}{{{\left( {{r}^{2}}-{{l}^{2}} \right)}^{2}}}=9\times {{10}^{9}}\times \frac{2\left( 500\times {{10}^{-6}}\times 0.1 \right)\times 0.25}{{{\left[ {{(0.25)}^{2}}-{{(0.05)}^{2}} \right]}^{2}}}$
$=6.25\times {{10}^{7}}~N/C(k=1$ for air $)$
NCERT Page-28 / N-24
Electric Charges and Fields
272186
Intensity of an electric field $\left( E \right)$ depends on distance $r$, due to a dipole, is related as
1 $E\propto \frac{1}{r}$
2 $E\propto \frac{1}{{{r}^{2}}}$
3 $E\propto \frac{1}{{{r}^{3}}}$
4 $E\propto \frac{1}{{{r}^{4}}}$
Explanation:
(c) Intensity of electric field due to a Dipole
$E=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}\sqrt{3co{{s}^{2}}\theta +1}\Rightarrow E\propto \frac{1}{{{r}^{3}}}$
272183
When an electric dipole $\vec{P}$ is placed in a uniform electric field $\vec{E}$ then at what angle between $\vec{P}$ and $\vec{E}$ the value of torque will be maximum?
1 ${{90}^{\circ }}$
2 ${{0}^{\circ }}$
3 ${{180}^{\circ }}$
4 ${{45}^{\circ }}$
Explanation:
(a)
NCERT Page-28 / N27
Electric Charges and Fields
272184
An electric dipole has a pair of equal and opposite point charges $q$ and $-q$ separated by a distance $2x$. The axis of the dipole is
1 from positive charge to negative charge
2 from negative charge to positive charge
3 perpendicular to the line joining the two charges drawn at the centre and pointing upward direction
4 perpendicular to the line joining the two charges drawn at the centre and pointing downward direction
Explanation:
(b)
NCERT Page-28 / N-27
Electric Charges and Fields
272185
The electric intensity due to a dipole of length $10~cm$ and having a charge of $500\mu C$, at a point on the axis at a distance $20~cm$ from one of the charges in air, is
1 $6.25\times {{10}^{7}}~N/C$
2 $9.28\times {{10}^{7}}~N/C$
3 $13.1\times {{10}^{11}}~N/C$
4 $20.5\times {{10}^{7}}~N/C$
Explanation:
(a) Given : Length of the dipole $\left( 2l \right)=10~cm=0.1~m$ or $l=$ $0.05~m$
Charge on the dipole $\left( q \right)=500\mu C=500\times {{10}^{-6}}C$ and distance of the point on the axis from the mid-point of the dipole $\left( r \right)=20+5=25~cm=0.25~m$.
We know that the electric field intensity due to dipole on the given point $\left( E \right)$
$=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2\left( q.2l \right)r}{{{\left( {{r}^{2}}-{{l}^{2}} \right)}^{2}}}=9\times {{10}^{9}}\times \frac{2\left( 500\times {{10}^{-6}}\times 0.1 \right)\times 0.25}{{{\left[ {{(0.25)}^{2}}-{{(0.05)}^{2}} \right]}^{2}}}$
$=6.25\times {{10}^{7}}~N/C(k=1$ for air $)$
NCERT Page-28 / N-24
Electric Charges and Fields
272186
Intensity of an electric field $\left( E \right)$ depends on distance $r$, due to a dipole, is related as
1 $E\propto \frac{1}{r}$
2 $E\propto \frac{1}{{{r}^{2}}}$
3 $E\propto \frac{1}{{{r}^{3}}}$
4 $E\propto \frac{1}{{{r}^{4}}}$
Explanation:
(c) Intensity of electric field due to a Dipole
$E=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}\sqrt{3co{{s}^{2}}\theta +1}\Rightarrow E\propto \frac{1}{{{r}^{3}}}$
272183
When an electric dipole $\vec{P}$ is placed in a uniform electric field $\vec{E}$ then at what angle between $\vec{P}$ and $\vec{E}$ the value of torque will be maximum?
1 ${{90}^{\circ }}$
2 ${{0}^{\circ }}$
3 ${{180}^{\circ }}$
4 ${{45}^{\circ }}$
Explanation:
(a)
NCERT Page-28 / N27
Electric Charges and Fields
272184
An electric dipole has a pair of equal and opposite point charges $q$ and $-q$ separated by a distance $2x$. The axis of the dipole is
1 from positive charge to negative charge
2 from negative charge to positive charge
3 perpendicular to the line joining the two charges drawn at the centre and pointing upward direction
4 perpendicular to the line joining the two charges drawn at the centre and pointing downward direction
Explanation:
(b)
NCERT Page-28 / N-27
Electric Charges and Fields
272185
The electric intensity due to a dipole of length $10~cm$ and having a charge of $500\mu C$, at a point on the axis at a distance $20~cm$ from one of the charges in air, is
1 $6.25\times {{10}^{7}}~N/C$
2 $9.28\times {{10}^{7}}~N/C$
3 $13.1\times {{10}^{11}}~N/C$
4 $20.5\times {{10}^{7}}~N/C$
Explanation:
(a) Given : Length of the dipole $\left( 2l \right)=10~cm=0.1~m$ or $l=$ $0.05~m$
Charge on the dipole $\left( q \right)=500\mu C=500\times {{10}^{-6}}C$ and distance of the point on the axis from the mid-point of the dipole $\left( r \right)=20+5=25~cm=0.25~m$.
We know that the electric field intensity due to dipole on the given point $\left( E \right)$
$=\frac{1}{4\pi {{\varepsilon }_{0}}}\times \frac{2\left( q.2l \right)r}{{{\left( {{r}^{2}}-{{l}^{2}} \right)}^{2}}}=9\times {{10}^{9}}\times \frac{2\left( 500\times {{10}^{-6}}\times 0.1 \right)\times 0.25}{{{\left[ {{(0.25)}^{2}}-{{(0.05)}^{2}} \right]}^{2}}}$
$=6.25\times {{10}^{7}}~N/C(k=1$ for air $)$
NCERT Page-28 / N-24
Electric Charges and Fields
272186
Intensity of an electric field $\left( E \right)$ depends on distance $r$, due to a dipole, is related as
1 $E\propto \frac{1}{r}$
2 $E\propto \frac{1}{{{r}^{2}}}$
3 $E\propto \frac{1}{{{r}^{3}}}$
4 $E\propto \frac{1}{{{r}^{4}}}$
Explanation:
(c) Intensity of electric field due to a Dipole
$E=\frac{p}{4\pi {{\varepsilon }_{0}}{{r}^{3}}}\sqrt{3co{{s}^{2}}\theta +1}\Rightarrow E\propto \frac{1}{{{r}^{3}}}$
